The Solution:
The question says two numbers are multiplied to get -50, but when added we get 5.
To find the two numbers, we shall say
Let one of the numbers be a, and the other b.
So that
![\begin{gathered} ab=-50\ldots eqn(1) \\ a+b=5\ldots eqn(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/whx15mat9qjdcvq11fh804xyzhwtnx7zvt.png)
Solving both equations simultaneously by the substitution method, we get
![\begin{gathered} \text{From eqn(2), we find a:} \\ a=5-b\ldots eqn(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/revuz3pdplj116lpw3lw5nn3dbwexthzn0.png)
Putting eqn(3) into eqn(1), we get
![\begin{gathered} (5-b)b=-50 \\ \text{Clearing the bracket, we get} \\ 5b-b^2=-50 \\ b^2-5b-50=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tmd0qnm7e24m2mqimed8a8ou53cmcayu0y.png)
Solving the above quadratic equation by the Factorisation Method, we get
![\begin{gathered} b^2-10b+5b-50=0 \\ b(b-10)+5(b-10)=0 \\ (b+5)(b-10)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hpq2fea4e6zoeo3q2o6bobwu1qlykqvhhq.png)
So,
![\begin{gathered} b+5=0\text{ or b-10=0} \\ b=-5\text{ or b=}10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6dacef62m9n11ln6h7agiu8v12pqnjspnv.png)
Substituting -5 or 10 for b in eqn(3), we get
![\begin{gathered} a=5--5=5+5=10 \\ or \\ a=5-10=-5 \\ So, \\ a=10,b=-5 \\ or \\ a=-5,b=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9uqf72rls0udxpo1aq7awcvh7yw3yxs8eg.png)
So, the two numbers are -5 and 10.
Therefore, the correct answers are -5 and 10